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7.1 Chemical equilibria and dynamic equilibrium

Syllabus
9701–2028–2029
Topic
7.1
Level
AS

Learning objectives

7.1.1What is meant by a reversible reaction• Reactions being equal and the concentration of reactants and products remaining constant- (a) understand what is meant by a reversible reaction- (b) understand what is meant by dynamic equilibrium in terms of the rate of forward and reverse- (c) understand the need for a closed system in order to establish dynamic equilibrium7.1.2Le Chatelier’s principle as: if a change• Define Le Chatelier’s principle as: if a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise this change7.1.3Le Chatelier’s principle to deduce (from• Use Le Chatelier’s principle to deduce qualitatively (from appropriate information) the effects of changes in temperature, concentration, pressure or presence of a catalyst on a system at equilibrium7.1.4Expressions for equilibrium constants• Deduce expressions for equilibrium constants in terms of concentrations, K c7.1.5The terms mole fraction and partial pressure• Use the terms mole fraction and partial pressure7.1.6Expressions for equilibrium constants• Deduce expressions for equilibrium constants in terms of partial pressures, K p (use of the relationship between Kp and Kc is not required)7.1.7The Kc and Kp expressions to carry out• Use the Kc and Kp expressions to carry out calculations (such calculations will not require the solving of quadratic equations)7.1.8The quantities present at equilibrium• Calculate the quantities present at equilibrium, given appropriate data7.1.9Whether changes in temperature, concentration• State whether changes in temperature, concentration or pressure or the presence of a catalyst affect the value of the equilibrium constant for a reaction7.1.10Conditions used in the Haber process• Describe/explain: the conditions used in the Haber process and the Contact process, as examples of the importance of an understanding of dynamic equilibrium in the chemical industry and the application of Le Chatelier’s principle

Dynamic equilibrium needs reversible reactions in a closed system

A reversible reaction can proceed from reactants to products and from products back to reactants under the stated conditions. It is represented by opposing half-arrows, ⇌.

At dynamic equilibrium, the forward and reverse reactions continue but have equal rates. Reactant and product concentrations therefore remain constant with time, even though particles keep reacting in both directions.

A closed system prevents reactants or products from escaping or being added. Without that material boundary, loss of a gaseous product or continuing feed/removal changes the composition and prevents the two rates from establishing a stable equilibrium state.

For N₂O₄(g) ⇌ 2NO₂(g) in a sealed vessel at constant temperature, the colour becomes constant when the two rates are equal. Individual N₂O₄ and NO₂ particles still interconvert.

Equal rates do not mean equal concentrations, and constant concentration does not mean the reaction has stopped. ‘Closed’ refers to no transfer of matter; it does not require the system to be thermally insulated.

Le Chatelier's principle predicts how equilibrium position responds

If a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise this change.

The position of equilibrium describes the relative equilibrium amounts of reactants and products. A move to the right produces more products and consumes some reactants; a move to the left does the reverse.

The response establishes a new dynamic equilibrium under the changed conditions. It reduces the imposed effect but does not normally restore every concentration, pressure or temperature to its original value.

The principle predicts the direction of composition change, not how fast equilibrium is reached or the numerical final amounts. Those applications and catalyst effects are handled by the next objective.

Classify the disturbance before predicting an equilibrium shift

Change imposed Direction favoured Boundary
add a reactant or remove a product forward, to the right responds to concentrations of species in the equilibrium
remove a reactant or add a product reverse, to the left the change is only partly opposed
increase pressure by decreasing volume side with fewer moles of gas no shift if gaseous mole totals are equal; ignore solids and liquids in the count
decrease pressure by increasing volume side with more moles of gas applies to gaseous equilibria
increase temperature endothermic direction identify the sign of the forward ΔH first
decrease temperature exothermic direction temperature changes both position and K
add a catalyst no shift both directions speed up, so equilibrium is reached sooner

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH < 0: adding O₂ shifts right; increasing pressure shifts right because three gas moles become two; increasing temperature shifts left because the reverse direction is endothermic; a catalyst does not change the final composition.

State how pressure was changed. Adding an inert gas at constant volume leaves reacting-gas partial pressures unchanged and does not shift the equilibrium, even though the total pressure rises.

Build Kc from equilibrium concentrations and equation coefficients

for aA+bB⇌cC+dD,Kc=[C]c[D]d[A]a[B]b\mathrm{for}\ aA+bB\rightleftharpoons cC+dD,\qquad K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}

Use equilibrium concentrations in mol dm⁻³. Put product terms above reactant terms and turn every stoichiometric coefficient in the balanced equation into a power.

N2+3H2⇌2NH3,Kc=[NH3]2[N2][H2]3\mathrm{N_2+3H_2\rightleftharpoons2NH_3},\qquad K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}

Include gases and dissolved species whose concentrations can change. Omit pure solids and pure liquids because their effective concentrations are constant; their presence in the balanced equation does not create a Kc term.

An equilibrium expression is derived from the equation as written. Reversing the equation reciprocates Kc and multiplying every coefficient by a factor raises Kc to that factor; never invent powers from measured concentrations.

Mole fraction converts a gas amount into its partial pressure

xi=nintotal∑xi=1x_i=\frac{n_i}{n_{total}}\qquad\sum x_i=1

pi=xiPtotal∑pi=Ptotalp_i=x_iP_{total}\qquad\sum p_i=P_{total}

The partial pressure of gas i is its contribution to the total pressure: for an ideal mixture it equals the pressure that the same amount of that gas would exert alone at the mixture's temperature and volume.

A mixture contains 2.00 mol N₂ and 1.00 mol O₂ at 300 kPa total pressure. x(N₂) = 2.00/3.00 = 0.667, so p(N₂) = 0.667 × 300 = 200 kPa; p(O₂) = 100 kPa, and the partial pressures sum to 300 kPa.

Mole fraction is dimensionless and is not mass fraction. Use all gaseous components in n(total), keep one pressure unit throughout, and verify both sums after calculating.

Build Kp using only equilibrium partial pressures of gases

for aA(g)+bB(g)⇌cC(g)+dD(g),Kp=pCcpDdpAapBb\mathrm{for}\ aA(g)+bB(g)\rightleftharpoons cC(g)+dD(g),\qquad K_p=\frac{p_C^c p_D^d}{p_A^a p_B^b}

Use each gas's equilibrium partial pressure, not the total pressure. Products go in the numerator, reactants in the denominator and balanced-equation coefficients become powers.

N2O4(g)⇌2NO2(g),Kp=pNO22pN2O4\mathrm{N_2O_4(g)\rightleftharpoons2NO_2(g)},\qquad K_p=\frac{p_{NO_2}^2}{p_{N_2O_4}}

Only gaseous species appear in Kp. Pure solids, pure liquids and aqueous concentrations are omitted from this partial-pressure expression. The unit, where requested, follows from the net pressure power in the particular expression.

Do not convert Kp into Kc: their relationship is outside this syllabus outcome. The task is to deduce Kp directly from the gas equation and the given partial pressures.

Calculate Kc or Kp only from equilibrium values

Write the expression before substituting, confirm that every value is an equilibrium concentration or partial pressure, apply the coefficient powers and derive the unit from the uncancelled concentration or pressure powers.

For N₂O₄(g) ⇌ 2NO₂(g):

Constant Equilibrium data Substitution Result
Kc [N₂O₄] = 0.200, [NO₂] = 0.300 mol dm⁻³ (0.300)² / 0.200 0.450 mol dm⁻³
Kp p(N₂O₄) = 80.0 kPa, p(NO₂) = 40.0 kPa (40.0)² / 80.0 20.0 kPa

The numerical result describes the equilibrium composition for this equation at the stated temperature. It does not measure reaction rate, and a Kc value cannot be substituted into a Kp expression.

Initial values are not equilibrium values unless the system initially happens to be at equilibrium. Keep brackets and powers intact, and do not report a unit copied from the input without cancelling the expression's powers.

Use stoichiometric change and K to recover every equilibrium quantity

Start from a balanced equation and one reaction extent x. Write initial, change and equilibrium amounts in the stoichiometric ratio, convert amounts to concentrations or partial pressures if required, substitute into K, solve the permitted non-quadratic relationship, then reject any value that gives a negative amount.

For H₂(g) + I₂(g) ⇌ 2HI(g) in 1.00 dm³, initially 1.00 mol H₂, 1.00 mol I₂ and no HI, with Kc = 49.0:

Stage / mol dm⁻³ H₂ I₂ HI
initial 1.00 1.00 0
change −x −x +2x
equilibrium 1.00−x 1.00−x 2x

49.0=(2x)2(1.00−x)2⇒7.00=2x1.00−x⇒x=0.777849.0=\frac{(2x)^2}{(1.00-x)^2}\Rightarrow 7.00=\frac{2x}{1.00-x}\Rightarrow x=0.7778

The equilibrium quantities are H₂ = 0.222 mol, I₂ = 0.222 mol and HI = 1.56 mol in the 1.00 dm³ vessel. Substitution returns Kc ≈ (1.56)²/(0.222)² = 49.0 after rounding.

Equal forward and reverse rates do not make equilibrium amounts equal. Apply coefficients to the change row—not independently guessed changes—and use the actual vessel volume whenever it is not 1.00 dm³.

Only temperature changes the equilibrium constant for a fixed reaction

Change Equilibrium position/composition Value of Kc or Kp
temperature changes; higher T favours the endothermic direction changes
concentration shifts until the equilibrium ratio is restored unchanged at constant T
pressure may shift a gaseous equilibrium unchanged at constant T
catalyst reaches the same equilibrium faster unchanged

For an exothermic forward reaction, increasing temperature shifts left and decreases K; decreasing temperature shifts right and increases K. For an endothermic forward reaction, these K changes reverse.

Concentration or pressure disturbances temporarily make the reaction ratio inconsistent with K, so composition changes until the same constant is restored. A catalyst lowers activation energy for both directions and changes neither the ratio nor K.

Always hold temperature constant when claiming that pressure or concentration leaves K unchanged. A pressure change caused by heating also changes temperature, so separate the variables before drawing the conclusion.

Haber and Contact conditions balance equilibrium yield, rate, cost and safety

Process and equilibrium Industrial conditions Equilibrium and rate explanation
Haber: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹ about 450 °C; about 200 atm; iron catalyst; NH₃ removed and unreacted gases recycled lower T and higher pressure favour NH₃, but very low T is slow and very high pressure is costly and hazardous; catalyst raises rate without changing yield
Contact: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH ≈ −197 kJ mol⁻¹ about 450 °C; near atmospheric pressure; V₂O₅ catalyst; oxygen available and SO₃ removed for conversion to sulfuric acid lower T and higher pressure favour SO₃, but 450 °C gives useful rate and the equilibrium yield is already high near 1 atm, so extra pressure is not worth its cost

Both forward reactions are exothermic and reduce gaseous mole count, so Le Chatelier predicts low temperature and high pressure for maximum equilibrium yield. Industry instead optimises production rate, energy and compression cost, safety, catalyst performance, separation and recycling.

Removing product shifts each operating equilibrium toward further product formation. In the Haber process ammonia is cooled and condensed; in the Contact process sulfur trioxide is removed from the gas stream and absorbed during sulfuric-acid manufacture.

A catalyst never increases equilibrium yield. The two processes also do not use the same pressure compromise: Haber gains enough yield and rate to justify high pressure, whereas Contact obtains little extra benefit from compression.